Can somebody please help me understand this Cauchy-Schwarz inequality? I am told that
$A\neq 0$ is Hermitian with nonzero eigenvalues $\lambda_{1},\cdots,\lambda_{r}$. And according to the Cauchy-Schwarz inequality it is obvious that…
$\newcommand{\tr}{\mathrm{tr}}(\tr (A))^{2}=(\Sigma_{i=1}^{k}(\lambda_{i}))^{2}\leq k\Sigma_{i=1}^{k}\lambda_{i}^{2}=k (\tr A^{2}) $
However, I do not follow this reasoning. Thanks in advance.
Best Answer
C-S inequality says $$|x_1.y_1+x_2.y_2+\cdots+x_r.y_r|^2\le|x_1^2+x_2^2+\cdots+x_r^2| \cdot |y_1^2+y_2^2+\cdots+y_r^2|$$
Now put $x_i=1$ and $y_i=\lambda_i$.